There is only one variable suitable for induction here, and that is $n$. The other variable $\epsilon$ does not range over the natural numbers, so just treat it like any other unspecified quantity.
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Harald Hanche-OlsenJan 25 '14 at 13:32

You've forgotten to show that $n\in\mathbb N$. Otherwise we couldn't prove the inequality using induction.
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user26486Jan 25 '14 at 13:39

1 Answer
1

Let's have a look at how much each side changes from $A(n)$ to $A(n+1)$. The left side is multiplied by $(1 + \varepsilon)$ in each step and the difference is therefore $\varepsilon(1 + \varepsilon)^n$. The difference of the right side is $2^{n+1} - 2^n = 2^n\varepsilon$.

The idea is now to compare the difference of both sides. If at the beginning the left side was less than or equal to the right side (and it was since $A(1)$ holds) and at the same time the difference between left and right side is not getting smaller, then the proposition holds for every $n$.

We will prove by induction that $A(n) \implies A(n+1)$. We can bound the difference of the left side with $\varepsilon\big(1 + (2^n - 1)\varepsilon\big) = \varepsilon + (2^n - 1)\varepsilon^2$, as we assume that $A(n)$ holds. We can further conclude that $$\varepsilon + (2^n - 1)\varepsilon^2 \le \varepsilon + (2^n - 1)\varepsilon,$$ as $\varepsilon \in [0,\ 1]$.